Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $n = \dfrac{8z + 40}{6} \times \dfrac{-2}{z(z + 5)} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $n = \dfrac{ (8z + 40) \times -2 } { 6 \times z(z + 5) } $ $ n = \dfrac {-2 \times 8(z + 5)} {6 \times z(z + 5)} $ $ n = \dfrac{-16(z + 5)}{6z(z + 5)} $ We can cancel the $z + 5$ so long as $z + 5 \neq 0$ Therefore $z \neq -5$ $n = \dfrac{-16 \cancel{(z + 5})}{6z \cancel{(z + 5)}} = -\dfrac{16}{6z} = -\dfrac{8}{3z} $